题解 | #三视图#

很好模拟题, 注意到真实三维坐标与数组的是反过来的, 也就是真实的越大, 越在上面, 但是数组的是越大, 越在下面, 因此需要将反置, 俯视图的情况和数组相同

#include <bits/stdc++.h>

#define x first
#define y second
#define all(x) x.begin(), x.end()

using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using PLD = pair<LD, LD>;

const int N = 1e5 + 10, MOD = 998244353;
const int INF = 1e9;
const LL LL_INF = 1e18;
const LD EPS = 1e-8;
const int dx4[] = {-1, 0, 1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};

istream &operator>>(istream &is, i128 &val) {
    string str;
    is >> str;
    val = 0;
    bool flag = false;
    if (str[0] == '-') flag = true, str = str.substr(1);
    for (char &c: str) val = val * 10 + c - '0';
    if (flag) val = -val;
    return is;
}

ostream &operator<<(ostream &os, i128 val) {
    if (val < 0) os << "-", val = -val;
    if (val > 9) os << val / 10;
    os << static_cast<char>(val % 10 + '0');
    return os;
}

struct Hash {
    vector<int> h, p;
    int B = 131;
    Hash (const string &s) {
        int n = s.size();
        h.resize(n + 1, 0);
        p.resize(n + 1, 1);
        for (int i = 0; i < n; ++i) {
            p[i + 1] = p[i] * B % MOD;
            h[i + 1] = (h[i] * B + s[i]) % MOD;
        }
    }

    LL get_hash(int l, int r) {
        LL v = h[r] - h[l - 1] * p[r - l + 1] % MOD;
        v = (v % MOD + MOD) % MOD;
        return v;
    }
};

bool cmp(LD a, LD b) {
    if (fabs(a - b) < EPS) return 1;
    return 0;
}

void solve() {
    int x, y, z, n;
    cin >> x >> y >> z >> n;
    vector<vector<int>> yx(y + 1, vector<int>(x + 1)), yz(y + 1, vector<int>(z + 1)), zx(z + 1, vector<int>(x + 1));
    for (int i = 0; i < n; ++i) {
        int a, b, c;
        cin >> a >> b >> c;
        yx[b][a] = 1;
        yz[b][c] = 1;
        zx[c][a] = 1;
    }

    // 这里需要将y反置
    for (int i = 1; i <= y; ++i) {
        for (int j = 1; j <= x; ++j) {
            cout << (yx[y - i + 1][j] ? 'x' : '.');
        }
        cout << ' ';
        for (int j = 1; j <= z; ++j) {
            cout << (yz[y - i + 1][j] ? 'x' : '.');
        }
        cout << '\n';
    }

    cout << '\n';

    for (int i = 1; i <= z; ++i) {
        for (int j = 1; j <= x; ++j) {
            cout << (zx[i][j] ? 'x' : '.');
        }
        cout << '\n';
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    while (T--) solve();
    cout << fixed << setprecision(15);

    return 0;
}